Source: Final Exam Practice Problems and Solutions
Tags: mass balance, material balance, conservation of mass, mixing, concentration, population balance, force vectors, resultant force, unit vectors, static equilibrium, free body diagram, ENGR 216, PHYS 216
Mass balance problems apply conservation of mass to track what enters and leaves a system: people in a town, salt in a mixture, solids in jam production. Force vector problems decompose each force into i-hat and j-hat components, then sum to find the resultant. Both topics lean on careful bookkeeping and getting signs right.
Mass balance (material balance)
Accumulation = Input − Output + Generation − Consumption. For non-reactive systems, generation and consumption are zero.
Concentration (mass fraction)
The proportion of a specific component in a mixture, often expressed as a percentage. Mass of component divided by total mass.
Resultant force
The single force vector that has the same effect as all the individual forces acting on a body, found by vector addition.
Unit vector
A vector of magnitude 1 that indicates direction. Standard 2D unit vectors are î (x-direction) and ĵ (y-direction).
Free body diagram (FBD)
A sketch isolating the body of interest with all external forces drawn as arrows at their points of application.
Static equilibrium
A body is in static equilibrium when the net force and net moment acting on it are both zero: ΣF = 0 and ΣM = 0.
The general balance: Final = Initial + In − Out + Born − Died
Booneville example:
1980 population: 3428
1990 population: 2783
Births: 178, Deaths: 87
2783 = 3428 + In − Out + 178 − 87
Net input (In − Out) = 2783 − 3428 − 178 + 87 = −736 people (net loss)
Realtors say 187 moved in, so: 187 − Out = −736, meaning Out = 923. 923 people left.
When mixing solutions, track the component of interest (salt, sugar, solids) separately.
Salt mixing example:
Solution A: 1.03 kg at 2.3% salt
Solution B: 3.04 kg at 4.8% salt
Total mass = 1.03 + 3.04 = 4.07 kg
Total salt = (0.023)(1.03) + (0.048)(3.04) = 0.02369 + 0.14592 = 0.16961 kg
Final concentration = 0.16961 / 4.07 = 4.17%
This is a more involved balance tracking water and solids through a process.
Regular jam (45:55 strawberry-to-sugar ratio):
Strawberries: 85% water, 15% solids, $0.50/lbm
Sugar: 100% solids, $0.10/lbm
Final jam: 66% solids
For 1.00 lbm of jam at 66% solids, you need 0.66 lbm solids.
In the initial mix (before heating), per 1 lbm: 0.45 lbm strawberries + 0.55 lbm sugar.
Solids from strawberries: 0.45 × 0.15 = 0.0675 lbm. Solids from sugar: 0.55 lbm.
Total solids per lbm mix: 0.6175 lbm. Water removed by heating until 66% solids.
To get 0.66 lbm solids in final jam: need (0.66/0.6175) = 1.069 lbm of initial mix.
Cost: (0.45 × 1.069 × $0.50) + (0.55 × 1.069 × $0.10) = $0.24 + $0.06 = $0.30
Sugar-free jam: Strawberries only (15% solids), target 66% solids.
Need 0.66 lbm solids, all from strawberries at 15% solids: 0.66/0.15 = 4.4 lbm strawberries.
Cost: 4.4 × $0.50 = $2.20
Every force in 2D can be written as F = Fₓ î + Fᵧ ĵ.
For a force at angle θ from the positive x-axis:
Fₓ = F cos(θ)
Fᵧ = F sin(θ)
For a force along a known direction given by a triangle (rise/run or two points):
Find the unit vector along that direction
Multiply the force magnitude by the unit vector
The resultant is found by decomposing each force into components and summing.
Resultant = −21 î − 40 ĵ (in lb)
The key steps:
For the cable tension (145 lb), find the direction from the geometry (84 in. horizontal, 80 in. vertical, cable length 116 in.)
For forces given by direction triangles (3-4-5 and 5-12-13), use the ratios directly
Sum all x-components, sum all y-components
Forces: 700 N at angle α from vertical, 300 N at 35° from vertical, 600 N at angle α from vertical.
Resultant = −38 î + 1295 ĵ (in N)
Given: Fₐ = 940 N, angles from diagram (50° and 70°).
Apply ΣFₓ = 0 and ΣFᵧ = 0 to solve for the two unknowns.
Results: F_B = 1020 N, F_C = 831 N
Cable geometry determines the angles at each segment. Apply equilibrium at each junction point.
Given: weight at B = 300 N.
Weight at C = 97.7 N
General mass balance (no reaction): Accumulation = In − Out
Mixing concentration: C_final = (m₁C₁ + m₂C₂) / (m₁ + m₂)
Force decomposition: F = F cos(θ) î + F sin(θ) ĵ
Unit vector from point P₁ to P₂: û = (P₂ − P₁) / |P₂ − P₁|
Resultant: R = ΣFₓ î + ΣFᵧ ĵ
⚠️ In mass balance problems, track the conserved quantity (salt, solids, people) separately from the total mass. Write two equations if needed.
⚠️ When decomposing forces, get the angle reference right. A force "50° from the vertical" has its x-component as F sin(50°), not F cos(50°).
⚠️ Direction triangles (e.g. 5-12-13) give you component ratios directly: no need to compute an angle first.
⚠️ In the jam problem, the heating step removes water but not solids. The mass of solids is conserved; only the total mass changes.
⚠️ For cable/equilibrium problems, draw the FBD at each connection point and write ΣFₓ = 0, ΣFᵧ = 0 at each.
Q: 1.03 kg of 2.3% salt solution is mixed with 3.04 kg of 4.8% salt solution. What is the final salt concentration?
A: Total salt = (0.023)(1.03) + (0.048)(3.04) = 0.1696 kg. Total mass = 4.07 kg. Concentration = 0.1696/4.07 = 4.17%.
Q: Why is sugar-free jam so much more expensive per pound than regular jam?
A: Strawberries are only 15% solids. To reach 66% solids without sugar, you need 4.4 lbm of strawberries at $0.50 each. Sugar (100% solids at $0.10/lbm) is a cheap way to add solids content.
Q: A force of 500 N acts along a direction defined by the 3-4-5 triangle. What are its components?
A: Fₓ = 500(3/5) = 300 N, Fᵧ = 500(4/5) = 400 N (signs depend on the actual direction in the diagram).
Q: In Booneville, if 187 moved in and the net input was −736, how many left?
A: 187 − Out = −736, so Out = 923.
mass balance, material balance, conservation of mass, mixing problems, concentration, mass fraction, salt solution, jam production, solids content, population balance, force vector, resultant, vector addition, component decomposition, unit vector, free body diagram, static equilibrium, cable tension, direction triangle, 3-4-5 triangle, ENGR 216 Module 6, ENGR 216 Module 7